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I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation

$$ A=B(\mathbb{1}-B)^{-1} $$

where B is a square matrix of dimension $N$ and $\mathbb{1}$ is $N \times N$ identity matrix.

Inspired by this post:

https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/

I was wondering if I really need to invert $\mathbb{1}-B$ in my case, or if there's some easier way. Keep in mind that:

  • $N$ in my case is quite large, its order of magnitude can be ten thousand.

  • I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_{00}$, $A_{01}$, $A_{11}$, $A_{02}$, $A_{12}$, $A_{22}$ would be perfect.

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Since $$ A = B(I-B)^{-1} = (I-B)^{-1}(I-B)B(I-B)^{-1} = (I-B)^{-1}B(I-B)(I-B)^{-1} =(I-B)^{-1}B $$ So you want to solve $$ (I-B)A=B $$ You seem to need only the first three columns of $A$. Solve the matrix problems $$ (I-B)a_i = b_i, \qquad i=0,1,2 $$ where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.

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